Question

What is the integral of `int sin^(4) (2x) dx`?

Answer 1

`int sin^4(2x) dx= (3x)/8-(sin(4x))/8 + (sin(8x))/64+C`

Explanation:

Use the trigonometric identities:

`sin^2 alpha = (1-cos2alpha)/2`

`cos^2 alpha = (1+cos2alpha)/2`

to have:

`sin^4(2x) = (sin^2(2x))^2 = (1-cos(4x))^2/4`

`sin^4(2x) = (1-2cos(4x) +cos^2(4x))/4`

`sin^4(2x) = 1/4-(cos(4x))/2 +1/8 +(cos(8x))/8`

`sin^4(2x) = 3/8-(cos(4x))/2 +(cos(8x))/8`

Using the linearity of the integral:

`int sin^4(2x) dx= 3/8int dx-1/2 int cos(4x)dx +1/8 intcos(8x)dx`

`int sin^4(2x) dx= (3x)/8-1/8 int cos(4x)d(4x) +1/64 intcos(8x)d(8x)`

`int sin^4(2x) dx= (3x)/8-(sin(4x))/8 + (sin(8x))/64+C`

Answer 2

The answer is `=1/64sin(8x)-1/8sin(4x)+3/8x+C`

Explanation:

Apply Euler's identity

`sin(2x)=(e^(2ix)-e^(-2ix))/(2i)`

`e^(ix)=cosx+isinx`

Therefore,

`sin^4(2x)=((e^(2ix)-e^(-2ix))/(2i))^4`

`=1/16((e^(i8x)+e^(-i8x))-4(e^(i4x)+e^(-i4x))+6)`

`=1/8((e^(i8x)+e^(-i8x))/2-4(e^(i4x)+e^(-i4x))/2+6/2)`

`=1/8cos(8x)-1/2cos(4x)+3/8`

So,

`intsin^4(2x)dx=int(1/8cos(8x)-1/2cos(4x)+3/8)dx`

`=1/64sin(8x)-1/8sin(4x)+3/8x+C`